5 mL solution of `H_2O_2` liberates 1.27 g of iodine from an acidified KI solution. What is the molarity of `H_2O_2`?

Explanation

`H_2O_2-=KI-=I_2`
`mEq-=mEq-=mEq`
`N_1xxV_1-=("weight of I_2")/(Ew(I_2))xx10^(3)mEq`
`N_1xx5mL-=(1.27xx10^(3))/((127xx2)/(2))` (n-factor of `I_2=2`)
`(2I^(ɵ)toI_2+2e^(-))`
Normality of `H_2O_2-=2N`
Molarity of `H_2O_2-=("normality")/(2)=(2)/(2)=1M`
(n-factor of `H_2O_2=2)(2e^(-)+H_2O_2to2H_2O)`